Problem: The altitude of the International Space Station $t$ minutes after its perigee (closest point), in kilometers, is given by $ A(t) = 415 - \sin\left(\dfrac{2\pi (t + 23.2)}{92.8}\right)$. The International Space Station reaches its perigee once in every orbit. How long does the International Space Station take to orbit the earth? Give an exact answer.
Explanation: The International Space Station reaches its perigee every time $A(t)$ reaches its minimum. $A(t)$ will reach its minimum whenever $\sin\left(\dfrac{2\pi (t+23.2)}{92.8}\right)$ reaches its maximum. That happens when $\dfrac{2\pi (t+23.2)}{92.8}$ is $\dfrac{\pi}{2}$ plus a multiple of $2\pi$. We can solve that equation: $\begin{aligned} \dfrac{2\pi (t+23.2)}{92.8} &= \frac{\pi}{2} + 2\pi n \\ \\ t + 23.2 &= 23.2 + 92.8 n \\ \\ t &= 92.8 n. \\ \end{aligned}$ The ISS is at its perigee whenever $t$ is a multiple of $92.8$ minutes. That happens every $92.8$ minutes. Since the ISS reaches its perigee once in every orbit, it takes $92.8$ minutes to orbit the earth.